Question

Resolve the fraction $\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}$ into partial fractions.

Answer 1

Hint: To convert $\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}$ into partial fractions, we will equate the fraction $\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}$ to $\dfrac{A}{\left( x-a \right)}+\dfrac{B}{\left( x+b \right)}$ where A and B are constants. Then we will solve the right side of this equation by taking LCM of the denominators of the two fractions. Then we will combine the coefficients of x and the constant terms. Then comparing these coefficients on both the sides of the equation, we will get two equations from which, we can find the value of A and B. From this, we can solve this question.

Complete step by step answer:
In the question, we are given a fraction $\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}$. We are required to convert this fraction into partial fractions.

Let us consider $\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{A}{\left( x-a \right)}+\dfrac{B}{\left( x+b \right)}$. Here A and B are constants.

Let us combine the two fractions on the right side by taking LCM.

$\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{A\left( x+b \right)+B\left( x-a \right)}{\left( x-a \right)\left( x+b \right)}$ . . . . . . . . . . . . (1)

$\Rightarrow \dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{Ax+Ab+Bx-Ba}{\left( x-a \right)\left( x+b \right)}$

Combining the coefficients of x and the constant terms, we get,

$\dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{x\left( A+B \right)+\left( Ab-Ba \right)}{\left( x-a \right)\left( x+b \right)}$

$\Rightarrow mx+n=x\left( A+B \right)+\left( Ab-Ba \right)$

Comparing the coefficient of x on both the sides, we get,

A + B = m . . . . . . . . (2)

Comparing the constant terms on both the sides, we get,

Ab – Ba = n . . . . . . . . . . (3)

Substituting B = m – A from equation (2) in equation (3), we get,

Ab – (m – A) a = n

$\Rightarrow $ A (a + b) = ma + n

$\Rightarrow A=\dfrac{ma+n}{a+b}$

Since B = m – A, we get,

$\begin{align}

  & B=m-\dfrac{ma+n}{a+b} \\

 & \Rightarrow B=\dfrac{ma+mb-ma-n}{a+b} \\

 & \Rightarrow B=\dfrac{mb-n}{a+b} \\

\end{align}$

So, we get,

$\begin{align}

  & \dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{\dfrac{ma+n}{a+b}}{\left( x-a \right)}+\dfrac{\dfrac{mb-n}{a+b}}{\left( x+b \right)} \\

 & \Rightarrow \dfrac{mx+n}{\left( x-a \right)\left( x+b \right)}=\dfrac{1}{\left( a+b \right)}\left( \dfrac{ma+n}{\left( x-a \right)}+\dfrac{mb-n}{\left( x+b \right)} \right) \\

\end{align}$

Hence, the answer is $\dfrac{1}{\left( a+b \right)}\left( \dfrac{ma+n}{\left( x-a \right)}+\dfrac{mb-n}{\left( x+b \right)} \right)$.


Note: To find the A and B, instead of comparing the coefficients, we can also substitute different values of x to find A and B. For example, if we substitute x = a in the equation (1), we get A (a+b) = am + n and hence, we get $A=\dfrac{ma+n}{a+b}$. Similarly, by substituting x = -b, we can find B.